Suppose we have a normally distributed population with a mea

Suppose we have a normally distributed population with a mean of 100 and a standard deviation of 20. if we take a sample size of 16 what is the probability that the mean of that sample size will be between 90 abd 110?

Solution

population mean = 100 and population s.d = 20..sample size =16..

let, X = mean of the sample size!

then, P [ 90 < X < 110 ]
=P [ ( 90 - 100) / ( 20 / SQRT ( 16) ) < ( X - mean) / std.error < ( 110-100) /( 20 / SQRT(16) ) ]
= P [ -2 < Z < 2 ] = P [ Z < 2 ] - P [ Z < -2 ] = 0.9544997...